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jeka94
2 years ago
5

Ammonia gas decomposes to form nitrogen and hydrogen gases. NH3(g) → N2(g) 3H2(g) If the nitrogen gas is collected in a rigid 2

liter metal cylinder at 22oC at 2000 kPa pressure, how many moles of nitrogen gas (N2) does the cylinder contain?.
Chemistry
1 answer:
sveticcg [70]2 years ago
6 0

Decomposition is a chemical reaction that breaks the reactant into two or more products. Moles of nitrogen gas (\rm N_{2}) in the cylinder is 1.63 moles.

<h3>What is the ideal gas equation?</h3>

The ideal gas equation states the relation of the hypothetical ideal gas according to the pressure, volume, temperature and moles of the gas. It is given by,

\rm PV = nRT

Where,

Pressure (P) = 2000 kPa

Volume (V) = 2L

Temperature (T) = 295 K

Gas constant (R)=  0.08206

Substituting values  in the equation:

\begin{aligned} \rm n &= \rm \dfrac{PV}{RT}\\\\&= \dfrac{2000 \times (\dfrac{1}{101.325}) \times 2}{0.08206 \times 295}\\\\&= 1.63\;\rm mol\end{aligned}

Therefore, 1.63 moles are produced.

Learn more about ideal gas equation here:

brainly.com/question/26720901

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A 20.00 ml sample of a solution of sr(oh)2 is titrated to the equivalence point with 40.03 ml of 0.1159 n hcl. what is the molar
goblinko [34]
The   molarity  of Sr(OH)2  solution is  =  0.1159 M

    calculation
write the equation  for reaction
that is,  Sr(OH)2 +2HCl→ SrCl2 + 2 H2O

then finds the mole  of HCl used

moles = molarity x volume 
=40.03 x0.1159 =  4.639 moles

by  use of mole ratio between Sr(OH)2 to  HCL which is 1 :2  the moles of Sr(OH)2  is therefore =  4.639  x1/2 = 2.312  moles

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Which of the following is true about the gas we call “air.” (the atmosphere)
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A mixture of hydrochloric and sulfuric acids is prepared so that it contains 0.315 M HCl and 0.125 M H2SO4. What volume of 0.55
Xelga [282]

<u>Answer:</u> The volume of NaOH required is 402.9 mL

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

  • <u>For HCl:</u>

Molarity of HCl solution = 0.315 M

Volume of solution = 503.4 mL = 0.5034 L   (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

0.315M=\frac{\text{Moles of HCl}}{0.5034L}\\\\\text{Moles of HCl}=(0.315mol/L\times 0.5034L)=0.1586mol

  • <u>For sulfuric acid:</u>

Molarity of sulfuric acid solution = 0.125 M

Volume of solution = 503.4 mL = 0.5034 L

Putting values in equation 1, we get:

0.125M=\frac{\text{Moles of }H_2SO_4}{0.5034L}\\\\\text{Moles of }H_2SO_4=(0.125mol/L\times 0.5034L)=0.0630mol

As, all of the acid is neutralized, so moles of NaOH = [0.1586 + 0.0630] moles = 0.2216 moles

Molarity of NaOH solution = 0.55 M

Moles of NaOH = 0.2216 moles

Putting values in equation 1, we get:

0.55M=\frac{0.2216}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.2216}{0.55}=0.4029L=402.9mL

Hence, the volume of NaOH required is 402.9 mL

6 0
3 years ago
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